All neighbor sum is 0 on a chessboard












10












$begingroup$


Neighbor cells of a matrix are defined as all the cells that share a common edge. A matrix is formed by putting real numbers in each cell.



Form an $8 times 8$ matrix (like a chessboard) where sum of all neighbors of each cell is $0$.




What is the fewest 0-valued cells possible after forming the matrix?







mathematics







share|improve this question











$endgroup$

















    10












    $begingroup$


    Neighbor cells of a matrix are defined as all the cells that share a common edge. A matrix is formed by putting real numbers in each cell.



    Form an $8 times 8$ matrix (like a chessboard) where sum of all neighbors of each cell is $0$.




    What is the fewest 0-valued cells possible after forming the matrix?







    mathematics







    share|improve this question











    $endgroup$















      10












      10








      10


      2



      $begingroup$


      Neighbor cells of a matrix are defined as all the cells that share a common edge. A matrix is formed by putting real numbers in each cell.



      Form an $8 times 8$ matrix (like a chessboard) where sum of all neighbors of each cell is $0$.




      What is the fewest 0-valued cells possible after forming the matrix?







      mathematics







      share|improve this question











      $endgroup$




      Neighbor cells of a matrix are defined as all the cells that share a common edge. A matrix is formed by putting real numbers in each cell.



      Form an $8 times 8$ matrix (like a chessboard) where sum of all neighbors of each cell is $0$.




      What is the fewest 0-valued cells possible after forming the matrix?







      mathematics




      mathematics






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Jan 9 at 20:56









      Sesquipedalian

      357




      357










      asked Jan 7 at 7:46









      OrayOray

      16.2k437157




      16.2k437157






















          2 Answers
          2






          active

          oldest

          votes


















          14












          $begingroup$

          This works for any given $a,b,c,d$:




          $$begin{array} {|c|c|c|c|c|c|c|c|}hline a&b&c&d&d&c&b&a\hline -b&-a-c&-b-d&-c-d&-c-d&-b-d&-a-c&-b\hline c&b+d&a+c+d&b+c+d&b+c+d&a+c+d&b+d&c\hline -d&-c-d&-b-c-d&-a-b-c-d&-a-b-c-d&-b-c-d&-c-d&-d\hline d&c+d&b+c+d&a+b+c+d&a+b+c+d&b+c+d&c+d&d\hline -c&-b-d&-a-c-d&-b-c-d&-b-c-d&-a-c-d&-b-d&-c\hline b&a+c&b+d&c+d&c+d&b+d&a+c&b\hline -a&-b&-c&-d&-d&-c&-b&-a\hline end{array}$$




          so:




          zero zeroes are needed (for example, $a,b,c,dgt0$)




          Following @Jaap's comment:




          $$begin{array} {|c|c|c|c|c|c|c|c|}hline a&h&c&g&d&f&b&e\hline -h&-a-c&-h-g&-c-d&-f-g&-b-d&-e-f&-b\hline c&g+h&a+c+d&f+g+h&b+c+d&e+f+h&b+d&f\hline -g&-c-d&-f-g-h&-a-b-c-d&-e-f-g-h&-b-c-d&-f-h&-d\hline d&f+g&b+c+d&e+f+g+h&a+b+c+d&f+g+h&c+d&g\hline -f&-b-d&-e-f-g&-b-c-d&-f-g-h&-a-c-d&-g-h&-c\hline b&e+f&b+d&f+h&c+d&g+h&a+c&h\hline -e&-b&-f&-d&-g&-c&-h&-a\hline end{array}$$







          share|improve this answer











          $endgroup$









          • 6




            $begingroup$
            You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
            $endgroup$
            – Jaap Scherphuis
            Jan 7 at 12:55



















          6












          $begingroup$

          Least amount of zero valued cells is




          zero




          One such possible grid is:




          enter image description here




          Because the grid size is even in both directions, I used the following algorithms to fill it:




          1. Fill 1 in first two cells. Some other number will also work.

          2. Copy the same number to the second end of row and additive inverse in opposite col (or vice versa).

          3. Fill rest of the number in a way that constraints are met.

          4. If any of the cells is zero, try with different numbers.




          I got the above algorithm as an intuition. I am still figuring an easy way to explain/prove why does it work?






          share|improve this answer











          $endgroup$














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            2 Answers
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            14












            $begingroup$

            This works for any given $a,b,c,d$:




            $$begin{array} {|c|c|c|c|c|c|c|c|}hline a&b&c&d&d&c&b&a\hline -b&-a-c&-b-d&-c-d&-c-d&-b-d&-a-c&-b\hline c&b+d&a+c+d&b+c+d&b+c+d&a+c+d&b+d&c\hline -d&-c-d&-b-c-d&-a-b-c-d&-a-b-c-d&-b-c-d&-c-d&-d\hline d&c+d&b+c+d&a+b+c+d&a+b+c+d&b+c+d&c+d&d\hline -c&-b-d&-a-c-d&-b-c-d&-b-c-d&-a-c-d&-b-d&-c\hline b&a+c&b+d&c+d&c+d&b+d&a+c&b\hline -a&-b&-c&-d&-d&-c&-b&-a\hline end{array}$$




            so:




            zero zeroes are needed (for example, $a,b,c,dgt0$)




            Following @Jaap's comment:




            $$begin{array} {|c|c|c|c|c|c|c|c|}hline a&h&c&g&d&f&b&e\hline -h&-a-c&-h-g&-c-d&-f-g&-b-d&-e-f&-b\hline c&g+h&a+c+d&f+g+h&b+c+d&e+f+h&b+d&f\hline -g&-c-d&-f-g-h&-a-b-c-d&-e-f-g-h&-b-c-d&-f-h&-d\hline d&f+g&b+c+d&e+f+g+h&a+b+c+d&f+g+h&c+d&g\hline -f&-b-d&-e-f-g&-b-c-d&-f-g-h&-a-c-d&-g-h&-c\hline b&e+f&b+d&f+h&c+d&g+h&a+c&h\hline -e&-b&-f&-d&-g&-c&-h&-a\hline end{array}$$







            share|improve this answer











            $endgroup$









            • 6




              $begingroup$
              You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
              $endgroup$
              – Jaap Scherphuis
              Jan 7 at 12:55
















            14












            $begingroup$

            This works for any given $a,b,c,d$:




            $$begin{array} {|c|c|c|c|c|c|c|c|}hline a&b&c&d&d&c&b&a\hline -b&-a-c&-b-d&-c-d&-c-d&-b-d&-a-c&-b\hline c&b+d&a+c+d&b+c+d&b+c+d&a+c+d&b+d&c\hline -d&-c-d&-b-c-d&-a-b-c-d&-a-b-c-d&-b-c-d&-c-d&-d\hline d&c+d&b+c+d&a+b+c+d&a+b+c+d&b+c+d&c+d&d\hline -c&-b-d&-a-c-d&-b-c-d&-b-c-d&-a-c-d&-b-d&-c\hline b&a+c&b+d&c+d&c+d&b+d&a+c&b\hline -a&-b&-c&-d&-d&-c&-b&-a\hline end{array}$$




            so:




            zero zeroes are needed (for example, $a,b,c,dgt0$)




            Following @Jaap's comment:




            $$begin{array} {|c|c|c|c|c|c|c|c|}hline a&h&c&g&d&f&b&e\hline -h&-a-c&-h-g&-c-d&-f-g&-b-d&-e-f&-b\hline c&g+h&a+c+d&f+g+h&b+c+d&e+f+h&b+d&f\hline -g&-c-d&-f-g-h&-a-b-c-d&-e-f-g-h&-b-c-d&-f-h&-d\hline d&f+g&b+c+d&e+f+g+h&a+b+c+d&f+g+h&c+d&g\hline -f&-b-d&-e-f-g&-b-c-d&-f-g-h&-a-c-d&-g-h&-c\hline b&e+f&b+d&f+h&c+d&g+h&a+c&h\hline -e&-b&-f&-d&-g&-c&-h&-a\hline end{array}$$







            share|improve this answer











            $endgroup$









            • 6




              $begingroup$
              You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
              $endgroup$
              – Jaap Scherphuis
              Jan 7 at 12:55














            14












            14








            14





            $begingroup$

            This works for any given $a,b,c,d$:




            $$begin{array} {|c|c|c|c|c|c|c|c|}hline a&b&c&d&d&c&b&a\hline -b&-a-c&-b-d&-c-d&-c-d&-b-d&-a-c&-b\hline c&b+d&a+c+d&b+c+d&b+c+d&a+c+d&b+d&c\hline -d&-c-d&-b-c-d&-a-b-c-d&-a-b-c-d&-b-c-d&-c-d&-d\hline d&c+d&b+c+d&a+b+c+d&a+b+c+d&b+c+d&c+d&d\hline -c&-b-d&-a-c-d&-b-c-d&-b-c-d&-a-c-d&-b-d&-c\hline b&a+c&b+d&c+d&c+d&b+d&a+c&b\hline -a&-b&-c&-d&-d&-c&-b&-a\hline end{array}$$




            so:




            zero zeroes are needed (for example, $a,b,c,dgt0$)




            Following @Jaap's comment:




            $$begin{array} {|c|c|c|c|c|c|c|c|}hline a&h&c&g&d&f&b&e\hline -h&-a-c&-h-g&-c-d&-f-g&-b-d&-e-f&-b\hline c&g+h&a+c+d&f+g+h&b+c+d&e+f+h&b+d&f\hline -g&-c-d&-f-g-h&-a-b-c-d&-e-f-g-h&-b-c-d&-f-h&-d\hline d&f+g&b+c+d&e+f+g+h&a+b+c+d&f+g+h&c+d&g\hline -f&-b-d&-e-f-g&-b-c-d&-f-g-h&-a-c-d&-g-h&-c\hline b&e+f&b+d&f+h&c+d&g+h&a+c&h\hline -e&-b&-f&-d&-g&-c&-h&-a\hline end{array}$$







            share|improve this answer











            $endgroup$



            This works for any given $a,b,c,d$:




            $$begin{array} {|c|c|c|c|c|c|c|c|}hline a&b&c&d&d&c&b&a\hline -b&-a-c&-b-d&-c-d&-c-d&-b-d&-a-c&-b\hline c&b+d&a+c+d&b+c+d&b+c+d&a+c+d&b+d&c\hline -d&-c-d&-b-c-d&-a-b-c-d&-a-b-c-d&-b-c-d&-c-d&-d\hline d&c+d&b+c+d&a+b+c+d&a+b+c+d&b+c+d&c+d&d\hline -c&-b-d&-a-c-d&-b-c-d&-b-c-d&-a-c-d&-b-d&-c\hline b&a+c&b+d&c+d&c+d&b+d&a+c&b\hline -a&-b&-c&-d&-d&-c&-b&-a\hline end{array}$$




            so:




            zero zeroes are needed (for example, $a,b,c,dgt0$)




            Following @Jaap's comment:




            $$begin{array} {|c|c|c|c|c|c|c|c|}hline a&h&c&g&d&f&b&e\hline -h&-a-c&-h-g&-c-d&-f-g&-b-d&-e-f&-b\hline c&g+h&a+c+d&f+g+h&b+c+d&e+f+h&b+d&f\hline -g&-c-d&-f-g-h&-a-b-c-d&-e-f-g-h&-b-c-d&-f-h&-d\hline d&f+g&b+c+d&e+f+g+h&a+b+c+d&f+g+h&c+d&g\hline -f&-b-d&-e-f-g&-b-c-d&-f-g-h&-a-c-d&-g-h&-c\hline b&e+f&b+d&f+h&c+d&g+h&a+c&h\hline -e&-b&-f&-d&-g&-c&-h&-a\hline end{array}$$








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Jan 7 at 13:09

























            answered Jan 7 at 11:03









            JonMark PerryJonMark Perry

            20.7k64099




            20.7k64099








            • 6




              $begingroup$
              You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
              $endgroup$
              – Jaap Scherphuis
              Jan 7 at 12:55














            • 6




              $begingroup$
              You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
              $endgroup$
              – Jaap Scherphuis
              Jan 7 at 12:55








            6




            6




            $begingroup$
            You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
            $endgroup$
            – Jaap Scherphuis
            Jan 7 at 12:55




            $begingroup$
            You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
            $endgroup$
            – Jaap Scherphuis
            Jan 7 at 12:55











            6












            $begingroup$

            Least amount of zero valued cells is




            zero




            One such possible grid is:




            enter image description here




            Because the grid size is even in both directions, I used the following algorithms to fill it:




            1. Fill 1 in first two cells. Some other number will also work.

            2. Copy the same number to the second end of row and additive inverse in opposite col (or vice versa).

            3. Fill rest of the number in a way that constraints are met.

            4. If any of the cells is zero, try with different numbers.




            I got the above algorithm as an intuition. I am still figuring an easy way to explain/prove why does it work?






            share|improve this answer











            $endgroup$


















              6












              $begingroup$

              Least amount of zero valued cells is




              zero




              One such possible grid is:




              enter image description here




              Because the grid size is even in both directions, I used the following algorithms to fill it:




              1. Fill 1 in first two cells. Some other number will also work.

              2. Copy the same number to the second end of row and additive inverse in opposite col (or vice versa).

              3. Fill rest of the number in a way that constraints are met.

              4. If any of the cells is zero, try with different numbers.




              I got the above algorithm as an intuition. I am still figuring an easy way to explain/prove why does it work?






              share|improve this answer











              $endgroup$
















                6












                6








                6





                $begingroup$

                Least amount of zero valued cells is




                zero




                One such possible grid is:




                enter image description here




                Because the grid size is even in both directions, I used the following algorithms to fill it:




                1. Fill 1 in first two cells. Some other number will also work.

                2. Copy the same number to the second end of row and additive inverse in opposite col (or vice versa).

                3. Fill rest of the number in a way that constraints are met.

                4. If any of the cells is zero, try with different numbers.




                I got the above algorithm as an intuition. I am still figuring an easy way to explain/prove why does it work?






                share|improve this answer











                $endgroup$



                Least amount of zero valued cells is




                zero




                One such possible grid is:




                enter image description here




                Because the grid size is even in both directions, I used the following algorithms to fill it:




                1. Fill 1 in first two cells. Some other number will also work.

                2. Copy the same number to the second end of row and additive inverse in opposite col (or vice versa).

                3. Fill rest of the number in a way that constraints are met.

                4. If any of the cells is zero, try with different numbers.




                I got the above algorithm as an intuition. I am still figuring an easy way to explain/prove why does it work?







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Jan 14 at 6:41

























                answered Jan 7 at 9:23









                Mohit JainMohit Jain

                2,8701842




                2,8701842






























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